CAIE P1 2015 June — Question 4 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeFind stationary points coordinates
DifficultyModerate -0.3 This is a straightforward optimization problem requiring substitution to eliminate one variable, expansion to get a quadratic in x, differentiation, and solving du/dx = 0. All steps are routine A-level techniques with no conceptual challenges, making it slightly easier than average but still requiring multiple standard procedures.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.07n Stationary points: find maxima, minima using derivatives

Variables \(u\), \(x\) and \(y\) are such that \(u = 2x(y - x)\) and \(x + 3y = 12\). Express \(u\) in terms of \(x\) and hence find the stationary value of \(u\). [5]

Question 4:
AnswerMarks
4u =2x(y−x) and x+3y =12,
12−x 
u =2x −x
 3 
=8x−8x2
3
du 16x
=8−
dx 3
= 0 when x = 11
2
→ ( y = 31 )
2
AnswerMarks
→ u = 6M1 A1
M1
A1
A1
AnswerMarks
[5]Expresses u in terms of x
Differentiate candidate’s quadratic,
sets to 0 + attempt to find x, or
other valid method
Complete method that leads to u
Co
Question 4:
4 | u =2x(y−x) and x+3y =12,
12−x 
u =2x −x
 3 
=8x−8x2
3
du 16x
=8−
dx 3
= 0 when x = 11
2
→ ( y = 31 )
2
→ u = 6 | M1 A1
M1
A1
A1
[5] | Expresses u in terms of x
Differentiate candidate’s quadratic,
sets to 0 + attempt to find x, or
other valid method
Complete method that leads to u
Co
Variables $u$, $x$ and $y$ are such that $u = 2x(y - x)$ and $x + 3y = 12$. Express $u$ in terms of $x$ and hence find the stationary value of $u$. [5]

\hfill \mbox{\textit{CAIE P1 2015 Q4 [5]}}